Hello
Does anyone know how to calculate the Ripple Voltage on a half-wave rectifier circuit and on a full-wave?
I found the following formulas but when I compare them with what I get on the oscilloscope, the difference is huge!

Originally posted by david mendes@Mar 23 2006, 02:03 PMHello
Does anyone know how to calculate the Ripple Voltage on a half-wave rectifier circuit and on a full-wave?
I found the following formulas but when I compare them with what I get on the oscilloscope, the difference is huge!

Ripple (half-wave)= 4,5 x I(mA)/C( μ F)

Ripple (full-wave)= 1,7 x I(mA)/C( μ F)

Cheers

David Mendes

[post=15349]Quoted post[/post]​

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The problem could be the constant or it could be the units that you are mixing. That is mA and uF. The approximations are derived from the following equation

Code ( (Unknown Language)):

dV/dT = - I/C

where

dV/dT is the voltage sag from one rectified peak to the next. dT will be 8.33 milliseconds for the full wave case and 16.67 milliseconds for the half wave case.

I is the current in Amperes

C is the capacitence in Farads.

In a less tha rigorous fashion this leads to

dV = dT*(I/C)

So using 0.00833 or 0.01667 for dT and I in amperes and C in Farads do you get more consistent results?

I was a little bit worried about the age of this thread, but i figured i'd try it anyway. Also thank you for replying. I'm just getting into some of my electrical engineering classes and this helped me out a lot. and a couple of my peers as well. thank you very much for sharing your knowledge so freely PapaBravo